Download CHAPTER Work and Energy

CHAPTER
6
Work and Energy
1* · True or false: (a) Only the net force acting on an object can do work. (b) No work is done on a particle that
remains at rest. (c) A force that is always perpendicular to the velocity of a particle never does work on the particle.
(a) False (b) True (c) True
2
· A heavy box is to be moved from the top of one table to the top of another table of the same height on the other
side of the room. Is work required to do this?
No; there is no change in potential or kinetic energy.
3
· To get out of bed in the morning, do you have to do work?
Yes, if your center of mass is higher when standing than when lying in bed. Otherwise, no.
4
· By what factor does the kinetic energy of a car change when its speed is doubled?
Note that K ∝ v2; K′ = 4K.
5* · An object moves in a circle at constant speed. Does the force that accounts for its acceleration do work on it?
Explain.
No; dW = F⋅d r and here F is perpendicular to d r.
6
· An object initially has kinetic energy K. The object then moves in the opposite direction with three times its
initial speed. What is the kinetic energy now? (a) K (b) 3K (c) -3K (d) 9K (e) -9K.
(d) K ∝ v2.
7
· A 15-g bullet has a speed of 1.2 km/s. (a) What is its kinetic energy in joules? (b) What is its kinetic energy if its
speed is halved? (c) What is its kinetic energy if its speed is doubled?
(a) K = 1/2mv2
K = 1/2(0.015)(1.2 × 103)2 J = 10.8 kJ
(b) K is reduced by a factor of 4
K = (10.8 kJ)/4 = 2.7 kJ
(c) K is increased by a factor of 4
K = 4(10.8 kJ) = 43.2 kJ
8
· Find the kinetic energy in joules of (a) a 0.145-kg baseball moving with a speed of 45 m/s and (b) a 60-kg
jogger running at a steady pace of 9 min/mi.
(a) K = 1/2mv2
K = 1/2(0.145)(45) 2 J = 147 J
(b) K = 1/2mv2; (1/9)mi/min=0.179km/min=2.98m/s
K = 1/2(60 kg))(2.98 m/s)2 = 267 J
Chapter 6
Work and Energy
9* · A 6-kg box is raised from rest a distance of 3 m by a vertical force of 80 N. Find (a) the work done by the
force, (b) the work done by gravity, and (c) the final kinetic energy of the box.
(a) W = Fy
W = 3 × 80 J = 240 J
(b) Wg = -mgy
Wg = -(6)(9.81)(3) J = -177 J
(c) K = W + Wg
K = 63 J
10 · A constant force of 80 N acts on a box of mass 5.0 kg that is moving in the direction of the applied force with a
speed of 20 m/s. Three seconds later the box is moving with a speed of 68 m/s. Determine the work done by this
force.
W = Kf - Ki = 1/2m(vf2 - vi2)
W = 1/2(5)[(68) 2 - (20) 2] J = 10.56 kJ
11 ·· You run a race with your girlfriend. At first you each have the same kinetic energy, but you find that she is beating
you. When you increase your speed by 25%, you are running at the same speed she is. If your mass is
85 kg, what is her mass?
2
2
m1 = 85 kg; solve for m2 = (85/1.56) kg = 54.4 kg
1/2m1v = 1/2m2(1.25v)
12 · How does the work required to stretch a spring 2 cm from its natural length compare with that required to stretch
it 1 cm from its natural length?
W = 1/2kx2; W ∝ x2; therefore work is increased by factor of 4.
13* ·· A 3-kg particle is moving with a speed of 2 m/s when it is at x = 0. It is subjected to a single force Fx that varies
with position as shown in Figure 6-30. (a) What is the kinetic energy of the particle when it is at x = 0?
(b) How much work is done by the force as the particle moves from x = 0 to x = 4 m? (c) What is the speed of
the particle when it is at x = 4 m?
(a) K(0) = 1/2mv2
K(0) = 1/2(3)(4) J = 6 J
(b) W = area under curve
W = 2 × 6 J = 12 J
2
2
(c) K(4) = K(0) + W = 1/2mv
1/2(3)v = 18 J; v = 3.46 m/s
14 ·· A 4-kg particle is initially at rest at x = 0. It is subjected to a single force Fx that varies with position as shown in
Figure 6-31. Find the work done by the force as the particle moves (a) from x = 0 to x = 3 m, and
(b) from x = 3 m to x = 6 m. Find the kinetic energy of the particle when it is at (c) x = 3 m and (d) x = 6 m.
(a) W = area under curve from x = 0 to x = 3
W(0-3) = 7.5 J
(b) See above but for x = 3 to x = 6
W(3-6) = -3 J
(c) Since K(0) = 0, K(3) = W(0-3), K(6) = W(0-6)
K(3) = 7.5 J; K(6) = 4.5 J
Chapter 6
Work and Energy
15 ·· A force Fx acts on a particle. The force is related to the position of the particle by the formula Fx = Cx3,
where C is a constant. Find the work done by this force on the particle when the particle moves from x = 1.5 m
to x = 3 m.
∫
W=
F dx
∫
W=C
3
1 .5
x3 dx =
C
4
(34 – 1.54) = 19 C J
16 ·· Lou’s latest invention, aimed at urban dog owners, is the X-R-Leash. It is made of a rubber-like material that
exerts a force Fx = -kx - ax2 when it is stretched a distance x, where k and a are constants. The ad claims,
“You’ll never go back to your old dog leash after you’ve had the thrill of an X-R-Leash experience. And you’ll
see a new look of respect in the eyes of your proud pooch.” Find the work done on a dog by the leash if the
person remains stationary and the dog bounds off, stretching the X-R-Leash from x = 0 to x = x0.
W=
∫ F dx
W=
∫
x0
0
1
2
(-kx – ax2)dx = - 2 kx 0 -
1
3
3
ax 0
17* ·· A 3-kg object is moving with a speed of 2.40 m/s in the x direction when it passes the origin. It is acted on by a
single force Fx that varies with x as shown in Figure 6-32. (a) What is the work done by the force from x = 0 to x = 2
m? (b) What is the kinetic energy of the object at x = 2 m? (c) What is the speed of the object at x = 2 m? (d) What is
the work done on the object from x = 0 to x = 4 m? (e) What is the speed of the object at x = 4 m?
(a) W = area under curve; count squares, each square
Between x = 0 and x = 2 there are about 22 squares,
= 0.125 J
corresponding to W = 2.75 J
(b) K = Ki + W
K = 1/2(3)(2.4) 2 J + 2.75 J = 11.4 J
(c) v = (2K/m)1/2
v = (2 × 11.4/3) 1/2 = 2.76 m/s
(d) As in (a) count squares
Net number of squares = 28; W = 3.5 J
(e) Proceed as in (b) and (c)
K = (8.64 + 3.5) J = 12.14 J; v = 2.84 m/s
18 ·· Near Margaret’s cabin is a 20-m water tower that attracts many birds during the summer months. During a hot
spell last year, the tower went dry, and Margaret had to have her water hauled in. She got lonesome without the birds
visiting, so she decided to carry some water up the tower to attract them back. Her bucket has a mass of 10 kg and
holds 30 kg of water when it is full. However, the bucket has a hole, and as Margaret climbed at a constant speed,
water leaked out at a constant rate. Several birds took advantage of the shower below, but when she got to the top,
only 10 kg of water remained for the birdbath. (a) Write an expression for the mass of the bucket plus water as a
function of the height y climbed. (b) Find the work done by Margaret on the bucket.
(a) Evidently, she loses 1 kg of water per meter
M(y) = (40 - y) kg
(b) W =
19 ·
∫
M(y)g dy
W = 9.81
∫
20
0
(40 – y)dy = 9.81[40 ×20 -
1
2
(20) 2] = 5886 J
Suppose there is a net force acting on a particle but it does no work. Can the particle be moving in a straight
Chapter 6
Work and Energy
line?
No; force is perpendicular to motion, therefore will cause departure from straight line motion.
20 · A 6-kg block slides down a frictionless incline making an angle of 60o with the horizontal. (a) List all the forces
acting on the block, and find the work done by each force when the block slides 2 m (measured along the incline). (b)
What is the total work done on the block? (c) What is the speed of the block after it has slid 1.5 m if it starts from
rest? (d) What is its speed after 1.5 m if it starts with an initial speed of 2 m/s?
(a) Forces acting are Fg = 6g down, FN = 6g cos 60o
Fnet = 6g sin 60o down along plane. W =12 × 9.81 ×
normal to plane
0.866 = 102 J = work done by Fg; FN does no work.
(b) Wtot = Wg
Wtot = 102 J
o
1/2
1/2
1/2
(c) v = (2K/m) = (2mg∆y/m) = (2g ∆y)
∆y = (1.5 sin 60 ) m; v = (2 × 9.81 × 1.5 × 0.866) 1/2
= 5.05 m/s
o
(d) a = g sin 60 ; use Equ. 2-15
v = (4 + 2 × 9.81 × 0.866 × 1.5) 1/2 m/s = 5.43 m/s
21* · An 85-kg cart is deposited on a 1.5-m platform after being rolled up an incline formed by a plank of length L that
has been laid from the lower level to the top of the platform. (Assume the rolling is equivalent to sliding without
friction.) (a) Find the force parallel to the incline needed to push the cart up without acceleration for
L = 3, 4, and 5 m. (b) Calculate directly from Equation 6-15 the work needed to push the cart up the incline for
each value of L. (c) Since the work found in (b) is the same for each value of L, what advantage, if any, is there
in choosing one length over another?
(a) F = mg sinθ = mg(1.5/L); mg = 834 N
L = 3 m, F = 417 N; L = 4 m, F = 313 N;
L = 5 m, F = 250 N
For L = 3 m, W = 3 × 417 J = 1.25 kJ; same for other L
(b) W = F.s
(b) Choosing longer length means one can exert a
smaller force
22 · A 2-kg object attached to a horizontal string moves with a speed of 2.5 m/s in a circle of radius 3 m on a
frictionless horizontal surface. (a) Find the tension in the string. (b) List the forces acting on the object, and find the
work done by each force during one revolution.
(a) T = mv2/r
T = 2(2.5) 2/3 N = 4.17 N
(b) Fg, FN, and T
None of the forces does any work.
What is the angle between the vectors A and B if A⋅B = -AB?
o
A⋅B = AB cos θ; cos θ = -1; θ = 180 .
23 ·
Two vectors A and B have magnitudes of 6 m and make an angle of 60o with each other. Find A⋅B.
o
2
2
A⋅B = (36 cos 60 ) m = 18 m .
24 ·
25* · Find A⋅B for the following vectors: (a) A = 3 i - 6 j, B = -4 i + 2 j; (b) A = 5 i + 5 j, B = 2 i - 4 j; and
(c) A = 6 i + 4 j, B = 4 i - 6 j.
(a), (b), (c) Use Equ. 6-12
(a) A⋅B = -24; (b) A⋅B = -10; (c) A⋅B = 0.
26 ·
Find the angles between the vectors A and B in Problem 25.
Chapter 6
Work and Energy
(a), (b), (c) θ = cos -1(A⋅B/AB)
(a) AB = 30,θ = 143o; (b) AB = 31.62,θ = 108o;
(c) θ = 90o
27 · A 2-kg object is given a displacement ∆s = (3 i + 3 j - 2 k ) m along a straight line. During the displacement, a constant force F = (2 i - 1 j + 1 k ) N acts on the object. (a) Find the work done by F for this displacement. (b) Find the
component of F in the direction of the displacement.
(a) W = F⋅∆s
W = (6 - 3 - 2) J = 1 J
(b) W = F⋅s cos θ = (F cos θ)∆s
F cos θ = (1 J)/(22 m2)1/2 = 0.213 N
28 ·· (a) Find the unit vector that is parallel to the vector A = Ax i + Ay j + Az k . (b) Find the component of the vector A
= 2 i - j - k in the direction of the vector B = 3 i + 4 j.
(a) The vector of unit length and parallel to A is uA = A/A.
(b) 1. Find the unit vector parallel to B
uB = (3/5) i + (4/5) j = 0.6 i + 0.8 j
2. The component of A along B is A⋅uB
A⋅uB = 1.2 + 0.8 = 2.0
29* ·· When a particle moves in a circle with constant speed, the magnitudes of its position vector and velocity vector are
constant. (a) Differentiate r⋅r = r2 = constant with respect to time to show that v⋅r = 0 and therefore v⊥r. (b)
Differentiate v⋅v = v2 = constant with respect to time and show that a ⋅v = 0 and therefore a ⊥v. What do the results of
(a) and (b) imply about the direction of a ? (c) Differentiate v⋅r = 0 with respect to time and show that a ⋅r + v2 = 0 and
therefore a r = -v2/r.
(a) (d/dt)(r⋅r) = r⋅(d r/dt) + (d r/dt)⋅r = 2v⋅r = 0. Therefore v⊥r.
(b) (d/dt)(v⋅v) = 2a ⋅v = 0. Therefore a ⊥v.
(c) The above implies that the component of a in the plane formed by r and v is colinear with r.
(d) (d/dt)(v⋅r) = v⋅(d r/dt) + r⋅(d v/dt) = v2 + r⋅a = 0. Therefore, a r = -v2/r.
30 ·· Vectors A, B, and C for a triangle as shown in Figure 6-33. The angle between A and B is θ, and the vectors are
related by C = A - B. Compute C⋅C in terms of A, B, and θ, and derive the law of cosines, C2 = A2 + B2 - 2AB cos θ.
1. Write C⋅C in terms of A and B
C⋅C = (A - B)⋅(A - B) = A⋅A + B⋅B - 2A⋅B
2. Write result in terms of magnitudes and θ
C2 = A2 + B2 - 2AB cos θ
31 · The dimension of power is (a) [M][L]2[T]2. (b) [M][L]2/[T]. (c) [M][L]2/[T]2. (d) [M][L]2/[T]3.
(d) P = F⋅v = [M][L/T 2][L/T] = [M][L]2/[T]3
32 · True or false: A kilowatt-hour is a unit of power.
False; it is a unit of energy.
33* ·
The engine of a car operates at constant power. The ratio of acceleration of the car at a speed of 60 km/h to that
at 30 km/h (neglecting air resistance) is (a)
1
2
. (b) 1/
2
. (c)
2
. (d) 2.
(a) mav = constant; av = constant and a ∝ 1/v .
34 ·· A car starts from rest and travels at constant acceleration. Which of the following statements are true? (a) The
power delivered by the engine is constant. (b) The power delivered by the engine increases as the car gains speed. (c)
The power delivered by the engine decreases as the car gains speed. (d) both (b) and (c) are correct.
Chapter 6
Work and Energy
(b) since v increases and a is constant, av must increase.
35 ·· Force A does 5 J of work in 10 s. Force B does 3 J of work in 5 s. Which force delivers greater power?
PA = (5/10) W = 0.5 W; PB = (3/5) W = 0.6 W; PB > PA.
36 · A 5-kg box is lifted by a force equal to the weight of the box. The box moves upward at a constant velocity of 2
m/s. (a) What is the power input of the force? (b) How much work is done by the force in 4 s?
(a) Find F; use Equ. 6-17
F = mg; P = Fv = mgv = (5 × 9.81 × 2) W = 98.1 W
(b) W = Pt
W = (98.1 × 4) J = 392.4 J
37* · Fluffy has just caught a mouse, and decides that the only decent thing to do is to bring it to the bedroom so that his
human roommate can admire it when she wakes up. A constant horizontal force of 3 N is enough to drag the mouse
across the rug at constant speed v. If Fluffy’s force does work at the rate of 6 W, (a) what is her speed, v? (b) How
much work does Fluffy do in 4 s?
(a) Use Equ. 6-17
v = (6/3) m/s = 2 m/s
(b) W = Pt
W = (6 × 4) J = 24 J
38 · A single force of 5 N acts in the x direction on an 8-kg object. (a) If the object starts from rest at x = 0 at time t =
0, find its velocity v as a function of time. (b) Write an expression for the power input as a function of time. (c) What
is the power input of the force at time t = 3 s?
(a) v = at; a = F/m
v = (5/8)t m/s = 0.625t m/s
(b) P = Fv
P = (5 × 0.625t) W = 3.125t W
(c) Substitute t = 3 s
P(3) = (3.125 × 3) W = 9.375 W
39 · Find the power input of a force F acting on a particle that moves with a velocity v for (a) F = 4 N i + 3 N k , v =
6 m/s i; (b) F = 6 N i - 5 N j, v = - 5 m/s i + 4 m/s j; and (c) F = 3 N i + 6 N j, v = 2 m/s i + 3 m/s j.
(a), (b), (c) Use P = F⋅v
(a) P = 24 W; (b) P = -50 W; (c) P = 24 W
40 ·· A particle of mass m moves from rest at t = 0 under the influence of a single force of magnitude F. Show that the
power delivered by the force at time t is P = F2t/m.
P = F⋅v; v = a t = (F/m)t; P = (F⋅F)(t/m) = F 2t/m.
41* ·· At a speed of 20 km/h, a 1200-kg car accelerates at 3 m/s2 using 20 kW of power. How much power must be
expended to accelerate the car at 2 m/s2 at a speed of 40 km/h?
P = Fv = mav
P = (1200 × 2 × 40/3.6) W = 26.7 kW
42 ·· A car manufacturer claims that his car can accelerate from rest to 100 km/h in 8 s. The car’s mass is 800 kg. (a)
Assuming that this performance is achieved at constant power, determine the power developed by the car’s engine.
(b) What is the car’s speed after 4 s? (Neglect friction and air resistance.)
(a) K = 1/2mv2 = Pt; solve for P = mv2/2t
P = [800 × (100/3.6) 2/16] = 38.6 kW
(b) K(4) = 1/2K(8); K ∝ v2
v(4) = v(8)/√ 2 = 70.7 km/h
43 ··
Show that the position of the truck in Example 6-11 is related to its speed by x = (m/3P)v3.
Chapter 6
Work and Energy
From Example 6-11, we have x = (8P/9m)1/2 t 3/2; from v = (2P/m)1/2 t1/2, t3/2 = (m/2P)3/2 v3. Substituting this
into the expression for x and simplifying, we obtain x = (m/3P)v3.
44 ·· A 700-kg car accelerates from rest under constant power. At the end of 8.0 s, its speed is 90 km/h and it is
located 133 m from its starting point. If the car continues to accelerate using the same power, what will its speed be at
the end of 10 s, and how far will the car be from the starting point?
From Example 6-11, v ∝ t1/2 and x ∝ t3/2
v(10) = v(8)(10/8) 1/2 = 100.6 km/h;
x(10) = x(8)(10/8) 3/2 = 186 m
45* ·· A 4.0-kg object initially at rest at x = 0 is accelerated at constant power of 8.0 W. At t = 9.0 s, it is at x = 36.0 m.
Find its speed at t = 6.0 s and its position at that instant.
v = (2Pt/m)1/2; x(6) = x(9)(6/9) 3/2
v = (2 × 8 × 6/4) 1/2 = 4.9 m/s; x(6) = (36)(6/9) 3/2 = 19.6 m
46 ·· A 700-kg car accelerates from rest under constant power at t = 0. At t = 9 s it is 117.7 m from its starting point
and its acceleration is then 1.09 m/s2. Find the power expended by the car’s engine, neglecting frictional losses.
From Example 6-11, P = 9mx2/8t3
P = [9 × 700 × (117.7) 2/8 × 93] W = 15 kW
47 · Two knowledge seekers decide to ascend a mountain. Sal chooses a short, steep trail, while Joe, who weighs the
same as Sal, goes up via a long, gently sloped trail. At the top, they get into an argument about who gained more
potential energy. Which of the following is true? (a) Sal gains more gravitational potential energy than Joe. (b) Sal
gains less gravitational potential energy than Joe. (c) Sal gains the same gravitational potential energy as Joe. (d) To
compare energies, we must know the height of the mountain. (e) To compare energies, we must know the length of
the two trails.
(c)
48 · The gravitational potential energy of an object changes by -6 J. It follows that the work done by the gravitational
force on this object is (a) - 6 J and the elevation of the object is increased. (b) -6 J and the elevation of the object is
decreased. (c) +6 J and the elevation of the object is increased. (d) +6 J and the elevation of the object is decreased.
(d)
49* · A woman runs up a flight of stairs. The gain in her gravitational potential energy is U. If she runs up the same
stairs with twice the speed, what will be her gain in potential energy? (a) U (b) 2U (c) U/2 (d) 4U (e) U/4
(a) The change in U does not depend on speed, only on difference in elevation.
50 · Which of the following statements is true? (a) The kinetic and potential energies of an object must always be
positive quantities. (b) The kinetic and potential energies of an object must always be negative quantities. (c) Kinetic
energy can be negative, but potential energy cannot. (d) Potential energy can be negative, but kinetic energy cannot.
(e) None of the preceding statements is true.
(d) U can be negative, depending on choice of zero; K = 1/2mv2 ≥ 0 since m and v2 are both positive or zero.
51 · A block slides a certain distance down an incline. The work done by gravity is W. What is the work done by
gravity if this block slides the same distance up the incline? (a) W (b) Zero (c) -W (d) Gravity can’t do work; some
other force does work. (e) Cannot be determined unless given the distance traveled.
(c) Force opposite to displacement, hence W < 0.
52 ·
True or false: (a) Only conservative forces can do work. (b) If only conservative forces act, the kinetic energy of
Chapter 6
Work and Energy
a particle does not change. (c) The work done by a conservative force equals the decrease in potential energy
associated with that force.
(a) False (b) False (for example, object in free fall) (c) True, by definition of U
53* · When you climb a mountain, is the work done on you by gravity different if you take a short, steep trail instead of a
long, gentle trail? If not, why do you find one trail easier?
No; a steeper trail requires more effort per step, but fewer steps.
54 · Which of the following forces are conservative and which are nonconservative? (a) the frictional force exerted on
a sliding box (b) the force exerted by a linear spring that obeys Hooke’s law (c) the force of gravity (d) the wind
resistance on a moving car
(a) nonconservative (b) conservative (c) conservative (d) nonconservative
55 ·
An 80-kg man climbs up a 6-m high flight of stairs. What is the increase in gravitational potential energy?
∆U = mgh
∆U = (80 × 6 × 9.81) J = 4.71 kJ
56 · One of the highlights of Sharika’s concert is her daredevil swan dive into the audience from a height of 2 m above
the crowd’s outstretched hands. If her mass is 60 kg, and the time of her dive is defined as t = 0, (a) what is her initial
potential energy relative to U = 0 at the position of the crowd’s hands? (b) From Newton’s laws, find the distance she
has fallen, and her speed at t = 0.20 s. (c) Find her potential and kinetic energy at t = 0.40 s. (d) Find her kinetic
energy and speed just as she reaches the hands of the crowd in the mosh pit.
(a) U = mgh
U = (60 × 2 × 9.81) J = 1.18 kJ
2
(b) v = gt; y = 1/2gt
v = (9.81 × 0.2) m/s = 1.96 m/s; y = 1/2(9.81 × 0.04) m
= 0.196 m
2
(c) y = 1/2gt ; ∆U = -mgy, and K = -∆U
y = 0.785 m; ∆U = -0.462 kJ, U = (1.18 - 0.462) kJ
= 0.718 kJ; K = 0.462 kJ
2
1/2
(d) K = -∆U = 1/2mv ; v = (2K/m)
K = 1.18 kJ; v = (2 × 1180/60) 1/2 m/s = 6.27 m/s
57* · Water flows over Victoria Falls, which is 128 m high, at an average rate of 1.4 × 106 kg/s. If half the potential
energy of this water were converted to electric energy, how much power would be produced by these falls?
P = -1/2(dU/dt) = -1/2gh(dm/dt)
P = 1/2(9.81 × 128 × 1.4 × 106) W = 879 MW
58 · A 2-kg box slides down a long, frictionless incline of angle 30o. It starts from rest at time t = 0 at the top of the
incline at a height of 20 m above ground. (a) What is the original potential energy of the box relative to the ground? (b)
From Newton’s laws, find the distance the box travels in 1 s and its speed at t = 1 s. (c) Find the potential energy and
the kinetic energy of the box at t = 1 s. (d) Find the kinetic energy and the speed of the box just as it reaches the
bottom of the incline.
(a) Ui = mgh
Ui = (2 × 9.81 × 20) J = 392 J
o
2
(b) a = g sin 30 ; x = 1/2at ; v = at
x = 1/2(9.81 × 0.5 × 1) m = 2.45 m; v = 4.91 m/s
2
(c) K = 1/2mv ; U = Ui - K
K = 1/2(2 × 4.912) J = 24.1 J; U = (392 - 24.1) J = 368 J
(d) K = Ui; v = (2K/m)1/2
K = 392 J; v = (2 × 392/2) 1/2 m/s = 19.8 m/s
59 ·
A force Fx = 6 N is constant. (a) Find the potential energy function U associated with this force for an
Chapter 6
Work and Energy
arbitrary reference position at x0 at which U = 0. (b) Find U such that U = 0 at x = 4 m. (c) Find U such that
U = 14 J at x = 6 m.
(a) From Equ. 6-21a, U(x) – U(x0) = -
∫
x
Fdx
U(x) = U(x0) -
x0
(b) U(x0) and x0 arbitrary
(c) See (b) above
∫
x
6dx = -6(x – x0) with U(x0) = 0
x0
U(x) = 24 - 6x
U(x) = 50 - 6x
60 · A spring has a force constant of k = 104 N/m. How far must it be stretched for its potential energy to be (a)
50 J and (b) 100 J?
(a), (b) U = 1/2kx2
(a) x = (2 × 50/104)1/2 m = 0.1 m; (b)x = (0.1 m) 2
= 0.141 m
61* ·· A simple Atwood’s machine uses two masses, m1 and m2 (Figure 6-34). Starting from rest, the speed of the two
masses is 4.0 m/s at the end of 3.0 s. At that instant, the kinetic energy of the system is 80 J and each mass has
moved a distance of 6.0 m. Determine the values of m1 and m2.
1. K = 1/2(m1 + m2)v2; solve for m1 + m2
m1 + m2 = (2 × 80/42) kg = 10 kg
2. a =
m1 − m 2
m1 + m 2
a = 1.33 m/s2; m1 - m2 = 1.33(10/9.81) kg = 1.36 kg
g = v(t)/t; solve for m1 - m2
m1 = 5.68 kg, m2 = 4.32 kg
3. Solve for m1 and m2
62 ·· A straight rod of negligible mass is mounted on a frictionless pivot as in Figure 6-35. Masses m1 and m2 are
suspended at distances l1 and l2. (a) Write an expression for the gravitational potential energy of the masses as a
function of the angle θ made by the rod and the horizontal. (b) For what angle θ is the potential energy a minimum? Is
the statement “systems tend to move toward a configuration of minimum potential energy” consistent with your result?
(c) Show that if m1l1 = m2l2, the potential energy is the same for all values of θ. (When this holds, the system will
balance at any angle θ. This result is known as Archimedes’ law of the lever.)
(a) Set U = 0 for θ = 0;
U(θ) = m2gl2 sin θ - m1gl1 sin θ = (m2l2 - m1l1)g sin θ
(b) Note that sin θ is maximum for θ = +π /2;
If m2l2 > m1l1 > 0, θ = -π /2 for Umin;
sin θ is minimum for θ = -π /2
if m2l2 < m1l1, θ = π /2 for Umin
This is consistent with the tendency for systems to move toward minimum potential energy.
(c) If m1l1 = m2l2, then (m2l2 - m1l1) = 0 and U is independent of θ.
63 ·· Figure 6-36 shows a plot of a potential-energy function U versus x. (a) At each point indicated, state whether the
force Fx is positive, negative, or zero. (b) At which point does the force have the greatest magnitude? (c) Identify any
equilibrium points, and state whether the equilibrium is stable or unstable.
(a) Use Fx = -dU/dx
A and E: Fx < 0; B, D and E: Fx = 0; C: Fx > 0
Chapter 6
(b) Find the point where slope is steepest
(c) Stable: d 2U/dx2 > 0; unstable: d 2U/dx2 < 0
Work and Energy
At point A Fx is greatest
B: unstable equilibrium; D: stable equilibrium
64 · (a) Find the force Fx associated with the potential-energy function U = Ax 4, where A is a constant. (b) At what
point(s) is the force zero?
(a) Fx = -dU/dx
Fx = -4Ax 3
(b) Set Fx = 0, solve for x
x=0
65* ·· A potential-energy function is given by U = C/x, where C is a positive constant. (a) Find the force Fx as a function
of x. (b) Is this force directed toward the origin or away from it? (c) Does the potential energy increase or decrease
as x increases? (d) Answer parts (b) and (c) where C is a negative constant.
(a), (b), (c) Fx = -dU/dx
(a) Fx = C/x2; (b) directed away from the origin
(-x direction); (c) U(x) increases with increasing x
(d) Repeat above with C < 0
Directed toward the origin; U increases as x increases
66 ·· On the potential-energy curve for U versus y shown in Figure 6-37, the segments AB and CD are straight lines.
Sketch a plot of the force Fy versus y.
Note that between A and B, F = +2 N; between C and
D, F = -2 N; between B and C, F changes gradually
from +2 N to -2 N.
67 ··
The force acting on an object is given by Fx = a/x 2. Determine the potential energy of the object as a function of x.
∫
∫
U(x) = - F(x)dx = - a/x2dx = a/x + U0.
68 ·· The potential energy of an object is given by U(x) = 3x2 - 2x3, where U is in joules and x is in meters. (a)
Determine the force acting on this object. (b) At what positions is this object in equilibrium? (c) Which of these
equilibrium positions are stable and which are unstable?
(a) F = -dU/dx
F = 6x2 - 6x = 6x(x - 1)
(b) At equilibrium, F = 0
Equilibrium at x = 0, x = 1
2
2
(c) d U/dx > 0 : stable equilibrium
d 2U/dx2 = 6 - 12x; stable at x = 0, unstable at x = 1
69* ·· During a Dead Wait concert, Sharika and Chico, each of mass M, are attached to the ends of a light rope that is
Chapter 6
Work and Energy
hung over two frictionless pegs, as shown in Figure 6-38. A large gong of mass m is attached to the middle of the rope,
between the pegs, and Sharika and Chico beat it madly in lieu of the usual guitar solo. (a) Find the potential energy of
the system as a function of the distance y shown in the figure. (b) Find the value of y for which the potential energy
function of the system is a minimum. (c) Find the equilibrium distance y0 using the potential energy function. (d) Check
your answer by applying Newton’s laws to the gong.
(a) Set U = 0 for y = 0; note that each M is raised by
U(y) = 2Mg[ (d 2 + y2)1/2 - d] - mgy
h = (d 2 + y2)1/2 - d as m drops a distance y
(b) Find dU/dy and set equal to zero
dU/dy = 2Mgy/(d 2 + y2)1/2 - mg; dU/dy = 0 for
(c) Set net force on m = 0
Solve for y
2
y = d(m/2M)/ 1 − ( m / 2m)
Fnet = mg - 2Mg sin θ; sin θ = y/(y2 + d 2)1/2
y = d(m/2M)/ 1 − (m / 2 M )
2
70 ··· The potential energy of an object is given by U(x) = 8x2 - x4, where U is in joules and x is in meters. (a)
Determine the force acting on this object. (b) At what positions is this object in equilibrium? (c) Which of these
equilibrium positions are stable and which are unstable?
(a), (b), (c) Proceed as in Problem 6-68
F = 4x3 - 16x = 4x(x + 2)(x - 2)
F = 0 at x = -2, 0, +2; these are equilibrium points.
d 2U/dx2 = 16 - 12x2 > 0 at x = 0, < 0 at x ±2
x = 0 stable , x = ±2 unstable equilibrium
71 ··· The force acting on an object is given by F(x) = x3 - 4x. Locate the positions of unstable and stable equilibrium and
show that at these points U(x) is a local maximum or minimum, respectively.
1. F(x) = x(x2 - 4); F(x) = 0 at x = 0, ±2.
At x = 0, a small increase in x makes F(x) negative, i.e., a
restoring force. Hence at x = 0, stable equilibrium.
At x = +2, an increase in x increases F(x); thus, unstable
equilibrium. Similarly, at x = -2, equilibrium is unstable.
4
2
2. U(x) = -x /4 + x /2
U(x) has local min. at x = 0, local max. at x = ±2
72 ··· The potential energy of a 4-kg object is given by U = 3x2 - x3 for x ≤ 3 m, and U = 0 for x ≥ 3 m, where U is in
joules and x is in meters. (a) At what positions is this object in equilibrium? (b) Sketch a plot of U versus x. (c)
Discuss the stability of the equilibrium for the values of x found in (a). (d) If the total energy of the particle is 12 J,
what is its speed at x = 2 m?
(a) Proceed as in Problem 6-68
dU/dx = 3x(2 - x) = 0 for x = 0, 2 m; also for x ≥ 3 m
(b) U(x) is shown on the right
Chapter 6
Work and Energy
(c) Note that U(x) is a minimum at x = 0 m; a
maximum at x = 2 m; and constant for x ≥ 3 m
(d) K = E - U; v = (2K/m)1/2
Equilibrium stable at x = 0 m; unstable at x = 2 m; neutral at
x≥3m
v = [2(12 - 4)/4] 1/2 m/s = 2 m/s
73* ··· A force is given by Fx = Ax -3, where A = 8 N.m3. (a) For positive values of x, does the potential energy associated
with this force increase or decrease with increasing x? (You can determine the answer to this question by imagining
what happens to a particle that is placed at rest at some point x and is then released.) (b) Find the
potential-energy function U associated with this force such that U approaches zero as x approaches infinity. (c)
Sketch U versus x.
(a) Apply Equ. 6-21a
U(x) = 1/2A/x 2 + U0; for x > 0, U decreases as x increases
(b) Let x → ∞ and set U(x) → 0
0 = 0 + U0; U0 = 0; U(x) = 1/2(8 N.m3)/x2 J = 4/x 2 J
(c) The plot of U(x) is shown
Chapter 6
Work and Energy
74 · True or false: (a) Only the net force acting on an object can do work. (b) Work is the area under the forceversus-time curve.
(a) False (b) False
75 · Negative work by an applied force implies that (a) the kinetic energy of the object increases. (b) the applied force
is variable. (c) the applied force is perpendicular to the displacement. (d) the applied force has a component that is
opposite to the displa cement. (e) nothing; there is no such thing as negative work.
(d) See Equ. 6-1.
76 ·· A movie crew is in the Badlands when their car overheats. After they stop to let it cool down, an argument breaks
out. They agree that they must go easy on the engine, but they disagree about when the engine works the hardest, and
therefore about how they should drive for the rest of the trip. Carolyn claims that the work done by the car in
accelerating from 0 to 20 km/h is less than that required to accelerate from 20 km/h to 30 km/h, meaning they should
drive more slowly. Ted says no, the work done between 0 and 20 km/h is more than the work done between 20 km/h
and 30 km/h. Ernie says it all depends on the mass of the car, and Bloop says it all depends on how long you take to
change from one speed to another. Who is right?
Note that W(0-20) = 1/2m × 400; W(20-30) = 1/2m(900 - 400) = 1/2m × 500; Carolyn is right.
77* · Figure 6-39 shows two pulleys arranged to help lift a heavy load. A rope runs around two massless, frictionless
pulleys and the weight w hangs from one pulley. You exert a force of magnitude F on the free end of the cord. (a) If
the weight is to move up a distance h, through what distance must the force move? (b) How much work is done on the
weight? (c) How much work do you do? (This is an example of a simple machine in which a small force F1 moves
through a large distance x1 in order to exert a large force F2 (= w) through a smaller distance x2 = h.)
(a) If w moves a distance h, F moves a distance 2h. (b) W = wh. (c) W = F × 2h; note that the tension, F, in the
string is 1/2w. Thus F × 2h = wh.
78 · In February 1995, a total of 54.3 billion kW-h of electrical energy was generated by nuclear power plants in the
United States. At the same time, the population of the United States was about 255 million people. If the average
American has a mass of 60 kg, and if the entire energy output of all nuclear power plants was diverted to supplying
energy for a single giant elevator, estimate the height h at which the entire population of the country could be lifted by
the elevator. In your calculations, assume that 25% of the energy goes into lifting the people; assume also that g is
Chapter 6
constant over the entire height h.
(NMgh) = 0.25E; h = 0.25E/NMg
Work and Energy
h=
(0.25 × 54.3 × 10 12 × 3600 )W
(2.55 × 10 8 × 60 × 9.81)N
= 3.26 × 105 m = 326 km
79 · One of the most powerful cranes in the world, operating in Switzerland, can slowly raise a load of M = 6000 tonne
to a height of h = 12.0 m (1 tonne = 1000 kg). (a) How much work is done by the crane? (b) If it takes 1.00 min to lift
the load at constant velocity to this height, find the power developed by the crane.
(a) W = mgh
W = (6 × 106 × 9.81 × 12) J = 7.06 × 108 J = 706 MJ
(b) P = W/t
P = (706/60) MW = 11.8 MW
80 · In Australia, there used to be a ski lift of length 5.6 km. It took about 60 min for the gondola to travel all the way
up. If there were 12 gondolas going up at once, each of mass 550 kg, and the angle of ascent was 30o, estimate the
power P of the engine needed to operate the ski lift.
P = NM ∆h/∆t
P = (12 × 550 × 5.6 × 103 × sin 30o)/60 W = 50.4 kW
81* · A 2.4-kg object attached to a horizontal string moves with constant speed in a circle of radius R on a frictionless
horizontal surface. The kinetic energy of the object is 90 J and the tension in the string is 360 N.
Find R.
2
2
R = (2 × 90/360) m = 0.5 m
1/2mv = K; mv /R = T; thus, R = 2K/T
82 · How high must an 800-kg Ford Escort be lifted to gain an amount of potential energy equal to the kinetic energy it
has when it is moving at 100 km/h?
2
2
h = [(100/3.6) 2/(2 × 9.81)] m = 39.3 m
1/2mv = mgh; h = v /2g
83 · The movie crew arrives in the Badlands ready to shoot a scene. The script calls for a car to crash into a vertical
rock face at 100 km/h. Unfortunately, the car won’t start, and there is no mechanic in sight. They are about to skulk
back to the studio to face the producer’s wrath when the cameraman gets an idea. They use a crane to lift the car by
its rear end and then drop it, filming at an angle that makes the car appear to be traveling horizontally. How high should
the 800-kg car be lifted so that it reaches a speed of 100 km/h in the fall?
From Problem 6-82, we see that h = 39.3 m.
84 ·· The force acting on a particle that is moving along the x axis is given by Fx = -ax2, where a is a constant.
Calculate the potential-energy function U relative to U = 0 at x = 0, and sketch a graph of U versus x.
Apply Equ. (6-21a)
U(x) = Fx dx = ax3/3 + U0
Find U0 by setting U(0) = 0
U0 = 0. U(x) = ax3/3
The plot of U(x) is shown
∫
Chapter 6
Work and Energy
85* ·· Water from behind a dam flows through a large turbine at a rate of 1.5 × 106 kg/min. The turbine is located 50 m
below the surface of the reservoir, and the water le aves the turbine with a speed of 5 m/s. (a) Neglecting any energy
dissipation, what is the power output of the turbine? (b) How many U.S. citizens would be supplied with energy by this
dam if each citizen uses 3 × 1011 J of energy per year?
(a) Einit/kg = gh = Efin/kg = 1/2vf2 + Wel/kg;
Pel = MWel/∆t
Pel = [1.5 × 106(9.81 × 50 - 1/2 × 52)/(60)] W = 12 MW
(b) 3 × 1011 J/y = 9.51 kW
N = 12 × 106/9.51 × 103 = 1261
86 ·· A force acts on a cart of mass m in such a way that the speed v of the cart increases with distance x as v = Cx,
where C is a constant. (a) Find the force acting on the cart as a function of position. (b) What is the work done by the
force in moving the cart from x = 0 to x = x1?
∫
(a) No change in U; thus E = F dx or F = dE/dx = d/dx(1/2mC2x2) = mC2x. (b) W = E = 1/2mC2x2.
87 ·· A force F = (2 N/m2)x2 i is applied to a particle. Find the work done on the particle as it moves a total
distance of 5 m (a) parallel to the y axis from point (2 m, 2 m) to point (2 m, 7 m) and (b) in a straight line from
(2 m, 2 m) to (5 m, 6 m).
(a) Since F is along x direction and displacement is along y direction W = ∫F⋅d s = 0.
(b) Evaluate ∫F⋅d s
∫ 2x dx = 3 x
5
W=
2
2
2
3
5
2 = 78 J
88 ·· A particle of mass m moves along the x axis. Its position varies with time according to x = 2t3 - 4t2, where x is in
meters and t is in seconds. (a) Find the velocity and acceleration of the particle at any time t; (b) the power delivered
to the particle at any time t; and (c) the work done by the force from t = 0 to t = t1.
(a) v = dx/dt; a = dv/dt = d 2x/dt2
v = (6t2 - 8t) m/s; a = (12t - 8) m/s2
(b) P = Fv = mav
P = 8mt(9t2 - 18t + 8) W
(c) W = 1/2m{[v(t1)]2 - [v(0)]2}
W = 2mt12(3t1 - 4)2
Chapter 6
Work and Energy
Note: W can also be obtained by integrating P(t) dt.
89* ·· A 3-kg particle starts from rest at x = 0 and moves under the influence of a single force Fx = 6 + 4x - 3x2, where
Fx is in newtons and x is in meters. (a) Find the work done by the force as the particle moves from x = 0 to x = 3 m.
(b) Find the power delivered to the particle when it is at x = 3 m.
3
(a) W = ∫Fx dx
3
W = 0 F(x)dx = (6x + 2x2 – x3) 2 = 9 J

∫
(b) Find v from v = (2K/m)1/2, K = W
Use P = Fv
v = (18/3) 1/2 m/s = 2.45 m/s
P = (-9 N)(2.45 m/s) = -22 W
90 ·· The initial kinetic energy imparted to a 20-g bullet is 1200 J. Neglecting air resistance, find the range of this
projectile when it is fired at an angle such that the range equals the maximum height attained.
Use Equ. 3-22 and the result of Problem 3-62
o
θ = 76 ;
R=
v 20
g
sin 2θ =
2 × 1200
sin 152° m = 5.74 km
0.02 × 9.81
91 ·· A force Fx acting on a particle is shown as a function of x in Figure 6-40. (a) From the graph, calculate the work
done by the force when the particle moves from x = 0 to the following values of x: -4, -3, -2, -1, 0, 1, 2, 3, and 4 m. (b)
Plot the potential energy U versus x for the range of values of x from -4 m to +4 m, assuming that U = 0 at x = 0.
(a) Work = area under force-versus-displacement curve. Note that for negative displacements, F is positive, so W
for x < 0 is negative. W(0,1) = 1 J; W(1,2) = -1 J; and for x > 2 m, W is negative.
Below is a tabulation of W for -4 m < x < 4 m.
x, m -4 -3 -2 -1 0 1 2 3 4
W, J -11 -10 -7 -3 0 1 0 -2 -3
(b) From Equ. 6-21a, ∆U = -W (U = 0 at x = 0 where
W = 0). The plot of U = ∆U = -W is shown.
92 ··
Repeat Problem 91 for the force Fx shown in Figure 6-41.
Chapter 6
Work and Energy
(a) Proceed as in Problem 6-91. The tabulation of x and W is below.
x, m -4 -3 -2 -1 0 1 2 3 4
W, J
6 4 2 1/2 0 1/2 3/2 5/2 3
(b) As before, U = -W. See the plot.
93* ·· A rope of length L and mass per unit length of u lies coiled on the floor. (a) What force F is required to hold one
end of the rope a distance y < L above the floor as shown in Figure 6-42? (b) Find the work required to lift one end of
the rope from the floor to a height l < L by integrating F dy from y = 0 to y = l.
∫
1
(a) The mass supported is µy; F = µyg. (b) W = 0 µ g y dy = µ g
l2
.
2
94 ··· A box of mass M is at the bottom of a frictionless inclined plane (Figure 6-43). The box is attached to a string that
pulls with a constant tension T. (a) Find the work done by the tension T when the box has moved a distance x along
the plane. (b) Find the speed of the box as a function of x and θ. (c) Determine the power produced by the tension in
the string as a function of x and θ.
(a) By definition, W = Tx. (b) Fnet = T - Mg sin θ; a = (T/M) - g sin θ; from Equ. 2-15,
v(x) = 2 (
T
− g sin θ ) x. . (c) P = Tv = T[2(T/M - g sin θ)x]1/2.
M
95 ··· A force in the xy plane is given by F = (F0/r)(yi - xj), where F0 is a constant and r =
x 2 + y 2 . (a) Show that
the magnitude of this force is F0 and that its direction is perpendicular to r = x i + y j. (b) Find the work done by this
force on a particle that moves in a circle of radius 5 m centered at the origin. Is this force conservative?
(a) F =(Fx2 + Fy2)1/2; If F⊥r, F⋅r = 0
Fx2 + Fy2 = (F02/r2)(y2 + x2) = F02;
(b) Since F⊥r, F is tangential to circle and constant.
At x = 5 m, y = 0, F points in the -j direction. If
d s is in -j direction (clockwise rotation), dW> 0.
F⋅r = (F0/r)(yi - xj)⋅(xi + yj) = (F0/r)(yx - xy) = 0
W = F0 × (2π r) = 10π F0 (clockwise rotation);
W = -10π F0 (counterclockwise rotation).
Not conservative; W ≠ 0 for complete circuit.
Chapter 6
Work and Energy
96 ··· A theoretical formula for the potential energy associated with the nuclear force between two protons, two
neutrons, or a neutron and a proton is the Yukawa potential:U = -U0(a/x)e-x/a, where U0 and a are constants. (a)
Sketch U versus x using U0 = 4 pJ (a picojoule, pJ, is 1 × 10-12 J) and a = 2.5 fm (a femtometer, fm, is 1 × 10-15 m). (b)
Find the force Fx. (c) Compare the magnitude of the force at the separation x = 2a to that at x = a. (d) Compare the
magnitude of the force at the separation x = 5a to that at x = a.
(a) A plot of U(x) is shown
(b) Fx = -(dU/dx)
(c) F(2a) = -U0(3/4a)e-2; F(a) = -U0(2/a)e-1
(d) F(5a) = -U0(6/25a)e-5
F(x) = -U0e-x/a[(a/x 2) + (1/x)]
F(2a)/F(a) = (3/8)e-1 = 0.138
F(5a)/F(a) = (6/50)e-4 = 0.0022